Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $q = \dfrac{t^2 - 7t + 12}{-4t + 12} \div \dfrac{3t - 12}{-9t - 81} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{t^2 - 7t + 12}{-4t + 12} \times \dfrac{-9t - 81}{3t - 12} $ First factor the quadratic. $q = \dfrac{(t - 4)(t - 3)}{-4t + 12} \times \dfrac{-9t - 81}{3t - 12} $ Then factor out any other terms. $q = \dfrac{(t - 4)(t - 3)}{-4(t - 3)} \times \dfrac{-9(t + 9)}{3(t - 4)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (t - 4)(t - 3) \times -9(t + 9) } { -4(t - 3) \times 3(t - 4) } $ $q = \dfrac{ -9(t - 4)(t - 3)(t + 9)}{ -12(t - 3)(t - 4)} $ Notice that $(t - 3)$ and $(t - 4)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -9\cancel{(t - 4)}(t - 3)(t + 9)}{ -12(t - 3)\cancel{(t - 4)}} $ We are dividing by $t - 4$ , so $t - 4 \neq 0$ Therefore, $t \neq 4$ $q = \dfrac{ -9\cancel{(t - 4)}\cancel{(t - 3)}(t + 9)}{ -12\cancel{(t - 3)}\cancel{(t - 4)}} $ We are dividing by $t - 3$ , so $t - 3 \neq 0$ Therefore, $t \neq 3$ $q = \dfrac{-9(t + 9)}{-12} $ $q = \dfrac{3(t + 9)}{4} ; \space t \neq 4 ; \space t \neq 3 $